Why do moving back and forth, and moving in a circle look the same on a graph of position versus time?

Physics

1 answer:

pentagon [3]3 years ago

3 0

Answer:

one dimension of motion on a circle is "back and forth"

Explanation:

Whether the position graphs look the same or not is a function of the acceleration (and velocity), and how position is measured.

For a circle centered at the origin, uniform motion around the circle will be equivalent to sinusoidal motion in the x- or y-directions. So, that motion is equivalent to sinusoidal motion "back and forth", however it may be generated.

The "back and forth" motions of a piston in a cylinder (connected to a crankshaft), and of a pendulum, are almost sinusoidal, but not quite. Their position graphs will differ slightly from the graph of position of an object moving around a circle.

On the other hand, if the circular motion is plotted as the length of the radius versus time, it will be a constant -- not "back and forth" at all.

In short, plots of similar motion will look similar.

You might be interested in

In an early attempt to understand atomic structure, Niels Bohr modeled the hydrogen atom as an electron in uniform circular moti

AleksandrR [38]

Answer:

The answer is below

Explanation:

Using Coulomb's law of electric field which is:

$F=k\frac{q_1q_1}{r^2}\\\\ k =constant=9*10^9\ Nm^2/C^2,q_1=q_2=1.6*10^{-19}C,r=5.29*10^{-11}m\\\\Substituting\ gives:\\\\F=9*10^9*\frac{(1.6*10^{-19})*(1.6*10^{-19})}{(5.29*10^{-11})^2} =8.22*10^{-8}N\\\\Both\ centripetal\ force\ is\ given\ by:\\\\F=m\frac{v^2}{r} \\\\m = mass\ of \ electron=9.11*10^{-31}g,v=speed\ of\ electron\\\\F=m\frac{v^2}{r} \\\\v=\sqrt{\frac{F*r}{m} } \\\\subsituting:\\\\v=\sqrt{\frac{8.22*10^{-8}*5.29*10^{-11}}{9.11*10^{-31}} } \\\\v=2.18*10^6\ m/s\\\\$

$But\ \omega=\frac{v}{r}=\frac{2.18*10^6}{5.29*10^{-11}} =4.13*10^{17}\\\\\omega=2\pi f; f=frequency\\\\f=\frac{\omega}{2\pi} =\frac{4.13*10^{17}}{2\pi} \\\\f=6.57*10^{15}\ Hz$

6 0

3 years ago

What must be the diameter of a cylindrical 120-m long metal wire if its resistance is to be 6.0 ω? the resistivity of this metal

Free_Kalibri [48]

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$ .

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ . Rearranging the above equation,